If the sum of series `1+(3)/(x)+(9)/(x^(2))+(27)/(x^(3))+ . . .. " to "oo` is a finite number, then

Options

A. `xlt3`
B. `xgt(1)/(3)`
C. `xlt(1)/(3)`
D. `xgt3`

Answer

To solve the problem, we need to analyze the series given:

The series is:
\[ S = 1 + \frac{3}{x} + \frac{9}{x^2} + \frac{27}{x^3} + \ldots \]

This series can be recognized as a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{3}{x} \).

### Step 1: Identify the conditions for convergence
For the sum of an infinite geometric series to be finite, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \]

### Step 2: Set up the inequality for the common ratio
Since \( r = \frac{3}{x} \), we set up the inequality:
\[ \frac{3}{x} < 1 \]

### Step 3: Solve the inequality
To solve the inequality \( \frac{3}{x} < 1 \):
1. Multiply both sides by \( x \) (assuming \( x > 0 \)):
\[ 3 < x \]
or
\[ x > 3 \]

### Step 4: Consider the case when \( x \) approaches 3
We also need to consider the case when \( x = 3 \):
- If \( x = 3 \), then \( r = \frac{3}{3} = 1 \), which means the series does not converge (the sum becomes infinite).

### Step 5: Conclusion
Thus, for the series to converge and be finite, \( x \) must be greater than 3:
\[ x > 3 \]

### Final Answer
The range of \( x \) for which the sum of the series is finite is:
\[ x > 3 \]